= Solution
The relevant boundary values are $a=\pi$ and $\sigma=0$, but the original complex <integral> defining $g(0)$ requires $\sigma>0$. Substituting zero into it is invalid: its real part has a nonintegrable $dx/(x\log x)$ singularity near zero. Its imaginary part is different, since for positive $\sigma$
$$
F(t,a,\sigma)=\int_0^\infty
\frac{x^{\sigma-1}e^{-tx}}{a^2+\log^2x}\,dx.
$$
This has a finite limit as $\sigma\downarrow0$. On $(0,1)$, $x^{\sigma-1}\le x^{-1}$, whose displayed logarithmic denominator makes the <integral> finite; at infinity, a fixed upper bound on $\sigma$ and exponential damping provide an integrable dominator. Thus <dominated convergence> gives $N(t)=\lim_{\sigma\downarrow0}F(t,\pi,\sigma)$.
For the asymptotics, take the imaginary-part parameter identity before passing to that limit, with a fixed $0<s<1/2$. Its endpoint factor converges to
$$
h_0(r)=\frac{\sin(\pi r)}\pi\Gamma(r)=\frac1{\Gamma(1-r)}
$$
by the <Gamma reflection formula>. Near zero, $\sin(\pi r)\Gamma(\sigma+r)$ is bounded uniformly for small $\sigma\ge0$, because the sine zero compensates the gamma pole. At $s>0$, the remainder <integral> itself remains absolutely convergent at $\sigma=0$ and is $O(t^{-s})$. Consequently
$$
N(t)=\int_0^s e^{-r\log t}\frac{dr}{\Gamma(1-r)}+O(t^{-s}).
$$
The endpoint function is analytic with value one, so <Watson's lemma> proves the <logarithmic expansion of Ramanujan's function>
$$
\boxed{N(t)\sim\sum_{n\ge0}\frac{c_n}{(\log t)^{n+1}},
\qquad\frac1{\Gamma(1-z)}=\sum_{n\ge0}c_n\frac{z^n}{n!}}.
$$
In particular, with $\gamma$ <Euler's constant>,
$$
\boxed{N(t)=\frac1{\log t}-\frac\gamma{(\log t)^2}
+\frac{\gamma^2-\pi^2/6}{(\log t)^3}
+O\bigl((\log t)^{-4}\bigr)}.
$$
This use of <imaginary-part regularization at a zero Mellin exponent> matters: the positive-$\sigma$ coefficient $\beta_0$ in part (b) is always zero, whereas the boundary endpoint function has $h_0(0)=1$. The fixed-positive-parameter expansion is not uniform as $\sigma\downarrow0$, so its coefficients cannot simply be sent to that boundary term by term.
Back to article page
